How do I factor 8x^3-18x31 completely?

2 Answers
Jul 26, 2015

Factor 8x^3 - 1

Explanation:

Algebraic identity: a^3 - b^3 = (a - b) (a^2 + ab + b^2)a3b3=(ab)(a2+ab+b2)

8x^3 - 1 = (2x - 1)(4x^2 + 2x + 1) 8x31=(2x1)(4x2+2x+1)

Jul 26, 2015

You use the formula for the difference of perfect cubes.

Explanation:

All you really have to do in order to factor this expression completely is use the formula for the difference of two perfect cubes

a^3 - b^3 = (a-b)(a^2 + ab + b^2)a3b3=(ab)(a2+ab+b2)

You can rewrite the original expression as

8x^3 - 1 = (2x)^3 - 1^38x31=(2x)313

This will get you

(2x)^3 - 1^3 = (2x - 1)[(2x)^2 + 2x * 1 + 1^2](2x)313=(2x1)[(2x)2+2x1+12]

(2x)^3 - 1^3 = (2x - 1) * (4x^2 + 2x + 1)(2x)313=(2x1)(4x2+2x+1)

The quadratic 4x^2 + 2x + 14x2+2x+1 cannot be factored further without using complex numbers, which I'm not sure you're supposed to use.